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makkiz [27]
3 years ago
13

Area of a rectangle with sides measuring 34.5 and 45.3 feet

Mathematics
2 answers:
MariettaO [177]3 years ago
6 0

Answer:

1,562.85

Step-by-step explanation:

34.5 times 45.3

Usimov [2.4K]3 years ago
5 0

Answer:

1562.85 ft^2

Step-by-step explanation:

area is L x W

plug in 34.5 and 45.3

34.5 x 45.3 = 1562.85

thus, our final answer is 1562.85

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I really need help at this. Pleas help!!
Alik [6]

Answer: x>-5

It is x is greater than or equal to -5

6 0
2 years ago
A null hypothesis is that the mean nose lengths of men and women are the same. The alternative hypothesis is that men have a lon
algol [13]

Answer:

b. There is not enough evidence to say that the populations of men and women have different mean nose lengths.

See explanation below.

Step-by-step explanation:

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different (men have longer mean nose length than women), the system of hypothesis would be:

Null hypothesis:\mu_{men} \leq \mu_{women}

Alternative hypothesis:\mu_{men} > \mu_{women}

Assuming that we know the population deviations for each group, for this case is better apply a z test to compare means, and the statistic is given by:

z=\frac{\bar X_{men}-\bar X_{women}}{\sqrt{\frac{\sigma^2_{men}}{n_{men}}+\frac{\sigma^2_{women}}{n_{women}}}} (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Let's assume that the calculated statistic is z_{calc}

Since is a right tailed test test the p value would be:

p_v =P(Z>z_{calc})=0.225

And we know that the p value is 0.225. If we select a significance level for example 0.05 or 0.1 we see that p_v >\alpha

And on this case we have enough evidence to FAIl to reject the null hypothesis that the means are equal. So then the best conclusion would be:

b. There is not enough evidence to say that the populations of men and women have different mean nose lengths.

8 0
3 years ago
17) In the following figure, ABCD is a rhombus. Find 'X'. Pls ans fastly
creativ13 [48]

Answer:

x = 3

Step-by-step explanation:

5x - 1 = 3x +5

-3x

2x - 1 = 5

+1

2x = 6

devide by x

x = 3

5 0
3 years ago
Use the method of completing the square to transform the quadratic equation into the equation form (x + p)2 = q.
Brilliant_brown [7]

\text{Use:}(*)\ (a+b)^2=a^2+2ab+b^2\\\\2x^2+2x+1=0\ \ \ \ |:2\\\\x^2+x+0.5=0\\\\x^2+2\cdot x\cdot0.5+0.5=0\\\\\underbrace{x^2+2\cdot x\cdot0.5+0.5^2}_{(*)}-0.5^2+0.5=0\\\\(x+0.5)^2-0.25+0.5=0\\\\(x+0.5)^2+0.25=0\ \ \ \ |-0.25\\\\(x+0.5)^2=-0.25\to D)

Answer: D) (x + 0.5)2 = -0.25

5 0
3 years ago
The temperature at noon in Los Angeles on a summer day was 88° F. During the day, the temperature varied from this by as much as
uysha [10]

Answer:

The range of possible temperatures is the interval [80.5°,95.5°]

Step-by-step explanation:

Let

x -----> possible temperatures for that day.

we know that

The absolute value inequality of the difference of the possible temperatures and the temperature at noon must be less than or equal to  7.5° F

so

\left|x-88\right|\le7.5

Solve the absolute-value inequality

<em>First case  (positive case)</em>

+(x-88) \leq 7.5

Adds 88 both sides

x \leq 7.5+88

x \leq 95.5\°

<em>Second case  (negative case)</em>

-(x-88) \leq 7.5

Multiply by -1 both sides

(x-88) \geq -7.5

Adds 88 both sides

x \geq -7.5+88

x \geq 80.5\°

therefore

The range of possible temperatures is the interval [80.5°,95.5°]

4 0
3 years ago
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